Ch. 5 & 6 Review handout

Ch. 5 & 6 Review handout - Chapter Five Summary Riemann...

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Unformatted text preview: Chapter Five Summary Riemann Sums Suppose is defined for a S CU S b. Let n be a positive integer. Subdivide [(1,1)] into n subintervals of length Ax = (I) ~ a)/n. The first subinterval extends from a to 9:1 : a + A23, the second from $1 to $2 = a + ZAx, the third from $2 to $3 = a + 3132:, and so on. For uniformity in notation, usually we denote m0 = a and atn = b. The jth subinterval extends from mj_1 = a + (j — 1)Am to In each subinterval7 choose a point. Let be the point in the jth subinterval. Corresponding to these Choices; form the Rieman sum ftxilAIL' + fixilAm + - ‘ ' + f($2)Aa?- In sigma notation, this is 77. Z f(:c;)A3:. i=1 Note that the Rieman sum is a number, being a sum of function values times lengths of subintervals. There are infinitely many different Rieman sums that can be formed for f on [a, b], corresponding to Choices of n and the points Example 1 Let flan) : :52 — x and let the interval be [~173]. Choose n = 5, Thus subdivide [—l,3] into five submtervals of length An: = 55112 2 The submtemals are [—11w1-l-4/5] : [—1,—1/5], [—1/5,3/5], [3/5,7/5], [7/5,11/5] and [11/5, 15/5]. Choose a point in each subinterval, for example, 931‘ : ~1/2,9:; : 0,x§=1,xf§= 9/5 and '= 3. The corresponding Riemann sum is : new = its? — my 1 3:1 3 was»<2>+<02—o><:>+<<1>2—1>e filer—2)<§>+<32—3>(§>=s Definite Integral It may happen that the Rieman sums 2;;1 f(:r;)A:r converge on some num— ber L as n is chosen larger Without bound, no matter how is chosen in [wj_l, 963-]. Then we call L the definite (Riemann) integral of f(:r) over [a,b], denoted f: f(m)da:. Properties of the Definite Integral g i g 2 é {2 fb cf(9:)dac 2 c f(:c)dx if c is constant (constants factor out of an integral) fflflx) + g(:r)]dm : f: + f:g(x)dm (integral of a sum is the sum of the integrals) w g($)]dx : fab f(:c)da: — f: g(m)dm (integral of a difference is the difference of the integrals) f: f(x)d:r = + f: f(:c)d2: (the integral from a to any point, plus the integral from that point to b, is the integral from a to b) I: = — f: f: = 0 If f(m) 2 0, then ff f(:c)d:c 2 0 1f9(w) : fer then finance 3 1: node If m 3 fix) g M, then m(bw a) g ff f(:c)da: g MU; — a) Evaluation of Definite Integrals If : f’(:r), then f(:n)d:r = F(b) — This is sometimes called the evaluation theorem, or the net. Change theorem. Often we write f: f(:1:)dx = Fm]; or f: mode» = [F(m)]’; Example 2 17/2 , / 3cos(a:)da; = 3sin(:r)]::?l21 : 3sin(7r/2) ~ 35in(7T/4) : 3 — 3g. 1r/4 Example 3 3 1 3 / (:33 ~ 4:0)da: : —3:4 — 2:52] 4 —1 —1 1 = [1534) — 2e?) — (El—D4 ~ 2(4)?) e 4. The Fundamental Theorem of Calculus If f is continuous, then dim melt : we). Example 4 d 1; a ~1ln(cos(t))dt = ln(cos(m)), ‘ and d .m E 3 @752 sec(t)dt :. em2 sec(m). The Fundamental Theorem and the Evaluation Theorem can be summarized as follows: I) f more = M) — F<a> if mm) = we) 2 l l l l l and Eda; 9” mdt = M)- The last expression can be generalized to d 9(1) 5 M ) mm = f(9(w))g’($) e f(h<m>>h'<m) Example 5 —d—/ sin(t2)dt=sin(:c2) d1} ,4 tan(m) i etdt = titan”) sec2(a:) (1117 2 and d 1:1(92) d 1 1 2 2 ~— 13 t: . m — . dz $2 cos() cos(n(m))m cos(q:-)( x) Substitution Method for Evaluating Integrals To evaluate f:f(g(x))g’(a:)das, let u = ThenI when x = a, u 2 57(0): and when a: : b, u : g(b), so (2 9(1)) x ’ .10 d2", = u du. [a f(9( M > (a) f( ) The method also applies to indefinite integrals: / f(g(x>>g’(sc)dz= / f(u)du. Example 6 Evaluate f$2(x3 + 5)9d$. Let u = 51:3 + 5, Then (in : 3m2d$, so $205910 2 §du and 1 1 l 2 3 9 _ 9,_ 10 __ 3 r10 /m(x +5)dx—3/udu—30u +0—3O(m +0) +0. Example 7 Evaluate [OW/2 eos(m) sin(sin(a:))d:r. Let u : Sin(a:). Then du :2 cos(as)dac. Further, when m = O, u = 5111(0) = 0, and when m = 7r/2, u = sin(7r/2)=1. Then M2 ' q' — 1' ——cosu1 /0 cos(x)sm(sm(m))dcc—/O s1n(u)du- ( )]O z w cos(1) + cos(0) : 1 —— cos(1). ; 1 Example 8 Evaluate f: mmdw. Let u 2111(95), so du : idm, Ifcc :2 e, u : ln(e) : 1. And if to 2 e4, u 2 ln(e4) = 4. Then /e4 1 /4 1 d /4 ~l/2d _————- : - u 2 u u e sax/haw?) 1 fl 1 4 : 2W2] 2 NZ » 2x/I = 2. 1 Integrals of Even and Odd Functions on Symmetric Intervals f(3:) is an even function on [—a,a] if f(—m) = flea), and flat) is an odd function if f(-—x) = For example7 1, m3, and sin(x) are odd functions on any interval [—a, (1],. while 302, :54, 005(50) and :02 COS(3:) are even on any [—a, a]. If f is odd on [—a,a], then f(:1:)dz : 0. If f is even on [—a,a], then [fa f(3:)dac 2 2f: a, 2 S E 9 g i Example 9 Evaluate f: $66—34 (3052(336) sin(x)dx. Because the function is odd on [w4, 4], we immediately have IE4 9666"“4 cos2 sin(a:)dm :— Integration by Parts If Mac) and are functions of 3:; then (as an indefinite integral), /udv 2 av ~ fwdu. In terms of the definite integral, b »/ab Udv : .— ula)v(a’) " / 'Udu. 0. Example 10 Evaluate fol flatly. V Write this as fol ye“2ydy. Let u = y and d'u = e‘2ydy, Then du = dy and v = A%e_29. Then 1 1 1 1 / y€_2ydy= [—Sye‘zy] ~/ *ée'Eydy 0 ~ 0_ 0 Example 11 Evaluate fln(a:)cl:r. We can let u = ln(a:) and do 2 dz. Then do = ids and v : ac, so /1n(gc)d:c : /udv =uv— /vdu = asln(:c) ~ (3) d2: = mm —,/dx = 231nm). ——x+C. .1? Integration by Partial Fractions We can sometimes evaluate f r(:c)dm, where r(m) is a rational function (qqu tient of polynomials), by writing as a sum of simpler quotients. This is called a partial fractions decomposition Suppose r(sc) = Egg—:3; a quotient of polynomials, and the degree of q(ac) exceeds the degree ofp(:c). The key lies in factoring the denominator polynomial We want to write Tm 730v) : —— : sum of simpler quotients, where we fill in the terms on the right as follows: (1) if (x — c) is a factor of q(ac), but (a: — c)2 is not; put in a term A m—«c' (2) If (an — c)2 is a factor, but — c)3 is not, put in two terms A mgr... xflc (w—c)?’ (3) If (:10 ~ c)3 is a factor, but (m * c)2 is not, put in three terms A + B + D cc—c (an—c)2 (z—c)3. The trend is apparent for linear factors. Do this for each linear factor, taking into account its power as it occurs in (4) If our? + In + c is a factor (without real roots), but (0.332 + 1930 + c)2 is not a factor of q(a:), put in a term A32+B a032+b03+o (5) If am2 + bx + c is a factor (without real roots), and (ax2 + bx + c)2 is also a factor of q(:c), but (azu'z + bar + c)3 is not, put in two terms A93+B Cx+D ——+_—. ax2+b33+c (am2+bx+c) Cf! i g i g l a § § 5 i i E i l The trend for irreducible factors am2 + bx + 0 follows that for linear factors. Repeat this inclusion of terms for each such quadratic factor of q(x), taking into accounts its power. The end result is Mac) written as a sum of "simpler" quotients, each of which we must then integrate. Example 12 Evaluate / -53: d 952 + 6x + 8 33‘ —5:c _ A + B m2+6x+8_$+4 m+2i For this to be true, we must have Write A(x+2) +B(m+4) : ——5m. Putting m = ~2, we get 2B : 10, so B =2 5. Putting :1: = —4, we get —2A 2 20, so A z —10. Then —5ac 5 10 m2+6m+8zx+2_x+4' “5x 5 10 _~_____d.: d _ /m2+6x+8 a /m+2 m _/m+4dx =51n|x+2|~101n|ac+41+C~ Then Example 13 Evaluate z2+2mw4 ~——~——-da. m3~3m2+m——3 First write x2+2w~4 x2+233—4 A Bac—l—C $3~3m2+x—3_(m—3)($2+1) x—3+$2+1' For this to be true, we must have A(x2+1)+(Bm+C)(m—3) :$2+2$~4. Then (A+B)a:2+(0~3B)m+A—BC:z2+2ze4. Equating coefi'icients of like powers ofm on both sides of this equation, we get A+B:1,C—SB=2 andA—3C=-—44 These equations have solution A = 11/10,B : ~1/10,C = 17/10. «mum; Then x2+203~4 _11 1 1 9: +17 1 m3—3m2+$~3_10x—3 10:132—1-1 10x2+1‘ Then 2724—232—4 11 1 2 17 /m3w3m2+m_3d$:mlni$“3l"%1n($ +1)+~1—6arctan(a§)+0. Numerical Approximation of Integrals There are many schemes for approximating integrals, including the midpoint rule, trapezoidal rule, Simpson’s rule7 and others. We will discuss only the midpoint rule, with error bound The midpoint rule is: n b __ a n / nodmbn Enronmzm—j) i=1 Where n is a positive integer; E7 is the midpoint of the jth interval, and 3 means "approximately equal”. The error EM in this approximation satisfies a — a>3 24n2 ’ Where K is chosen so that [f”(x)| 3 K for a g :1: g b. This error bound is used to determine how to choose n to make the approximation as accurate as is wanted, [EMISK 1/2 . Example 14 Approm'mate 0 Sln($2)d$, with an error not exceeding 1/100 (or 0.01). Let = sin(m2). Then f’(m) : 2:5 cos($2) and f”(x) : ZCOS($2) — 4$251n($2). For 0 g :r 5 1/2, certainly lf”(a:)| S 2 + 4(1/2)2 = 3, so we may choose K :2 3 in the error estimate. Further, a 2 0 and b 2 1/2, 30 b— a = 1/2. To get the approximation within 1/100, choose it so that — 3 1 2 3 Ka a):3</>:1< 1_ 24712 247i2 64712 100 This will be true if > 199 64 or 100 71 > "64* —— Thus choose n = 2, the smallest integer exceeding 1.25. Then A3: = 95f : %. Subdioide [0,1/2] into two subintervals of length 1/4. These subintervals are [0,1/4] and [1/4,1/2]. The midpoints are, respectively, 1/8 and 3/8. The midpoint approximation is 1/2 1 /0 sin(x2)dm :3 a [sin((1/8)2)+ sin((3/8)2))] : [llama/64) + sin(9/64)) z 0038947, This approximation is accurate to within 1/100, We could get better accuracy by choosing n larger. For example, to get an approximation accurate to within 1/1000000; we would choose n so that (b— a)3 * 3(1/2)3 _ 1 1 K _. _ < . 24712 24n2 64n2 1000000 Now we would need n2 > W, so we could choose n z 125. This would involve a lot more calculation. Improper Integrals An integral of the form f(a:)dm, over an infinite interval, is one type of integral called an improper integrali It is natural to think of this integral in terms of the limit: 00 T / f(a:)d:1: = lim / f(m)da:. If the limit on the right is a number L, we say that faoo is a convergent integral, and give it the value L: f(x)dx = L, If the limit is infinite or does not exist, we say that is a divergent integral, and do not give it a numerical value. Example 15 Consider flee idm. [fr 2 1, then f: idol: : ln(r), and limrhoolnfi) = 00. Therefore floo idcc diverges and has no numerical value. ' Example 16 Consider vNow = 1—% and li1nr_,(,o (1 —- = 1. Therefore is a convergent improper integral and we write :0 fidcc : 1, There is a comparison test for improper integrals which is useful when we cannot perform the integration explicitly. Suppose 0 g g(a:) g for r 2 (L. Then (1) If [:0 f(:c)dm converges, so does g(m)d:c. (2) If g(m)dz diverges, so does faoo f(:z:)dsc. However; if g(:c)da: converges, or if f(w)dm diverges, no conclusion can be drawn. Notice that this test does not tell us what the integral converges to7 even when it shows that the integral converges Example 17 Test I100 e_$2d9: for convergence. We cannot integrate e'm2d$ in simple form to take the limit as r ~4> 00. However, if‘z 2 1, then But flee e_”d:r converges. This can be determined by direct integration as fol- . ' 1’ _ _ . _ 00 _ lows. First, fl 6 mda: : e — e T, so limrfiafie — e T) : e, and fl e $d$ converges (to e). By (I) of the comparison test, I100 e_12d:r converges also (al- though probably not to e). We can also consider improper integrals of the form fix, f(:r)d:r as lirnMwoo f: f(a:)da:. Example 18 [30° exdr converges. To see this, integrate ff exda: : 63 ~ er. Since lirnT__,__OO(e3 — er) 2 ‘e3, then fix, emda: converges to e3, and we write [300 emdzr 2 63. Finally, we can consider improper integrals ff; f (flair by writing such an integral as ffoo f(9c)da: +1?) f(a:)dx, with c any number. For If; f(m)d:r to converge, both of these improper integrals must converge. ()0 0 _oo emdrc : fix) exdcr+ fox ewdm. It is easy to check that [MOO ezdx converges to 1, but that few emdic , 00 i diverges, so VLDO ewdm diverges. Example 19 ff; emda: diverges. To see this, first write f Chapter Six Summary Area Between Two Curves Suppose a region R is bounded above by the graph of and below by the graph 0fg(ac) for a S :c S I). Then b area of R 2/ -— g(ar))dx. Note of caution: in this expression, always subtract the lower boundary function from the upper boundary function. In some cases, the curves may cross and interchange roles as upper and lower boundary. In this case, break up the region into as many regions as necessary and write the area as a sum of areas, in each of which the correct upper and lower boundary is identified. Example 20 Find the area of the region in the first quadrant lying between the graphs of'g = 2 — 9:2 and y = $122. The graphs and the region whose area we want are shown. The graphs in— tersect at (1,1). For 0 g (r g 1, the upper boundary is part of the graph of y = 2 — 332 and the lower boundary is part of the graph ofy : :32. But for 1 S x g \/—2—, the upper boundary is part of the graph ofy : m2 and the lower boundary is part of the graph ofy : 2 — :02. The graphs have "crossed over ” at their point of intersection. Therefore, 1 x/i area 2/ [(2 fl 5132) — £82] dac +/ [£82 —— (2 —— x2)]d:r 0 1 2 1 2 “5 : [23: w fits] + [—13 —— 2.13] 3 0 3 1 3 8 2 —— é- —- 3 2. Volume of a Solid by the Method of Slices Suppose we want to compute the volume of some solid. This requires some information about the solid. Assume that it is possible to draw a. straight line through the solid (call this line the m— axis) in such a way that the cross section through the solid perpendicular to this line at at has known area A(:c), for a g a: g h. Then H 03le + col» 5 | [\3 E I ([0 + [\3 b volume of the solid 2 / A(:e)d$. Example 21 An elliptical region S in the gig—plane is bounded by the graph of 9mg + 41/2 = 36. A solid has S as base, and the cross section perpendicular to the a:- azis at :18 is an isoscles right triangle with hypotenuse in. the base 5. Find the volume of this solid. All we need is the area of the cross section at m. This section (see diagram) is a right triangle, having hypotenuse of length 3V4 ~ x2. If the other two sides have length L, then by the Pythagorean theorem, 2L2 : 9(4 w 372), so = %\/4 — :62. The area of this triangle is (basexheight) = ; (1W) A”) Z n n NIH The volume of the solid is 2 9 volume 2 / —(4 — $2)d:r -2 4 2 9 13 9 8 8 =~ ~—' =~ ——- 8—— =2. 4[4m 3z]fl2 4(8 3+ 3) 4 Are Length The length of the graph of y : fix), for a g at g b, is b 3 length 2/ x/1+ [f’(:t)]2dm. E 10 Example 22 Find the length of the graph ofy : m3 for 1 S :t g 2. Since f’(m) = 39:2, this length is 2 length : f V 1 + 9z4dz. 1 As often happens with integrals for arc length, this has no elementary evaluation. However, using the midpoint rule with n = 16 subdivisions of [1, 2], we get 2 length : / V 1 + 9m4da: :1: 7.8014. 1 Computation of Work Sometimes we want to compute the work done by a variable force in moving an object from one point to another. Often we approach such a problem by thinking of the quantity being moved as subdivided into layers (by putting an axis through the quantity and subdividing). We then approximate the forces involved on each layer, as well as the distance this layer must be moved, and approximate the work done in moving this layer as the product of the force times the distance. We then sum these quantities over the layers (forming a Riemann sum) and take the limit as the layers are made thinner, forming a Riemann integral for the work to be computed. Example 23 A circular pool with a flat bottom has radius 24 feet, and sides 5 feet high. The depth of the water is 4 feet. I Find the work done in pumping the water out of the tank. Imagine the pool as a circular cylinder five feet high, with radius twenty four feet. Put an .r— ascis down the axis of the cylinder, as in the diagram. The water fills the pool for 0 S at S 4. Subdivide the water into thin cylinders parallel to the bottom of the pool, and compute the work done in lifting each cylinder to the top of the pool. To do this, subdivide [0,4] into n subinteruals. The ith subinterual is [ca-,4, Assuming that the density of the water is 62.5 pounds per cubic foot, then the mass of the cylinder of water lying ibetween 33,11 and :0,- is 62.577(242)(33,——xi_1), which is the volume of this cylinder of water, multiplied by the density of water. Acceleration due to gravity is 32 ft/secg, so by Newton’s law the force this cylinder of water exerts down is approximately 62.57r(32)(24)2(x1 — $.11). The cylinder of water must be lifted to the top of the pool, a distance of approx— imately 5 ~ m2- feet. The work this will take is approximately 62.5r(32)(24)2(5 - mm — sci—1). which has the dimensions offorce times distance. The work done in lifting each cylinder of water is approximately Z 62.577(32)(24)2(5 — axial —- mm). i=1 11 g. E l E i 3 i 3 l m: In the limit as n —) 00, this Riemann sum approaches the actual work clone in lifting all of the water to the top of the pool (where it magically disappears), But this Riemann sum also approaches the integral f; 62.57T(32)(24)2(5 — :r)dm. Therefore 4 work done = / 62.5n(32)(24)2(5 — 3:)dm. 0 Now 332 4 4 /(5—m)dm=5mu—] :12 0 2 0 30 work clone : 12(62.5)(32)(24)27r = 1.382400% foot pounds. Supplement to problems listed in the syllabus: page 4-21, 17 - 20 (using only the midpoint approximation method) page 458, 32 — 37. 12 r mwmm ,r'rnmsuza-afi‘r: svl " )‘3 :35» ans ...
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This note was uploaded on 09/26/2008 for the course MA 126 taught by Professor O'neil during the Summer '06 term at University of Alabama at Birmingham.

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Ch. 5 & 6 Review handout - Chapter Five Summary Riemann...

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